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    • Thread Starter

    With substitution of u = 2 - x^2:

    INT [ x / (2 - x^2) ] dx

    My method:
    u = 2-x^2 therefore, x = ROOT(2-u)

    INT [ ROOT(2-u) / u ] dx to change dx to du
    INT [ { ROOT(2-u) / u } du/dx ] dx

    du/dx = -2x = -2ROOT(2-u)

    INT [ -2*ROOT(2-u)*ROOT(2-u) / u ] du
    INT [ -2/u ] du
    = -2 ln(u) = WTF?

    I cant follow the mark scheme :/

    well the the start, since u= 2- x^2, du/dx= -2x.

    why dont you use this fact to get rid of the x at the top instead of messing with roots?
    • Thread Starter

    Omg its just clicked: a simple idea that makes it SOOOO much easier.

    When converting dx --> du, look at your du/dx equation and pretend du/dx is a fraction....

    du/dx = -2x
    dx = du/-2x.

    Then substitue that BACK in as your DX. Wow SO much easier than my method.

    Ok, So when takling as substitution question, change dx to du first, then see whats left. Easy as hell now. Thank you

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